Lipschitzian solutions of the implicit Cauchy problem \(g(x')=f(t,x)\), \(x(0)=0\), with \(f\) discontinuous in \(x\).
Let a be a positive real number, n a positive integer, \(S=[0,a]\times {\mathbb{R}}^ n\). Denote by \({\mathcal I}\) the family of all subsets of S for which at least one of the projections on the axes is a set of Lebesgue measure zero. Moreover, let \({\mathcal G}\) be the family of all subsets Y of \({\mathbb{R}}^ n\) such that, for each \(i=1,2,...,n\), the supremum and the infimum of the projection of \(\overline{conv}(Y)\) on the ith axis are both positive or both negative. Our main result is the following theorem. Let Y be a non-empty arcwise connected subset of \({\mathbb{R}}^ n\) belonging to \({\mathcal G}\); g a continuous real function on Y; f a real function on S; A a subset of S belonging to \({\mathcal F}\). Assume that: (i) for each \(r\in {\mathbb{R}}\), the boundary of \(f^{-1}(r)\) in S belongs to \({\mathcal F}\); (ii) the restriction of f to \(S\setminus A\) is continuous and bounded and the closure in \({\mathbb{R}}\) of its range is contained in the range of g. Under such hypotheses, there exist a set \(E\subseteq [0,a]\), of measure zero, and a Lipschitzian function h: [0,a]\(\to {\mathbb{R}}^ n\) such that \(g(h'(t))=f(t,h(t))\) a.e. in [0,a], \(h(0)=0\), and the restriction of h' to [0,a]\(\setminus E\) is continuous. Several counterexamples show that none of the hypotheses of the theorem above can be dropped. In particular, assumption (i) cannot be replaced with the more general one: ''for each \(r\in {\mathbb{R}}\), the boundary of \(f^{-1}(r)\) in S is a set of Lebesgue measure zero in \({\mathbb{R}}^{n+1}\)''. To see this, an example is presented.
scientific article; zbMATH DE number 3925294